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Homework Help: Astronaut at centre of tetrahedron

  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data
    An astronaut in the International Space-station attaches himself to the four vertices of a regular tetrahedron shaped frame with 4 springs. The mass of the springs and their rest length are negligible, their spring constants are ##D_1=150 N/m##, ##D_2=250 N/m##, ##D_3=300 N/m## and ##D_4=400 N/m##. For the sake of simplicity - the astronaut is considered pointlike, its mass is m=70 kg. What is the period of the oscillation of the astronaut if he is displaced from his equilibrium position and released?

    2. Relevant equations

    3. The attempt at a solution
    I haven't yet attempted the problem by writing down the equations when the astronaut is displaced from centre because looking at the answer given, I think there is a shorter way. The given answer is

    $$2\pi \sqrt{\frac{m}{D_1+D_2+D_3+D_4}}$$​
    Looking at the answer, it is similar to the time period of mass hanging by four parallel springs. Is this by coincidence or is there a reason behind this? I am clueless if there is a reason.

    If it is a coincidence, I am ready to make the equations.

    Any help is appreciated. Thanks!
  2. jcsd
  3. Jan 9, 2014 #2
    Direction of displacement isn't given? :confused:
  4. Jan 9, 2014 #3
    I checked the problem again. I have copied the problem word by word. :)
  5. Jan 9, 2014 #4
    I can't think of any clever way.
  6. Jan 9, 2014 #5

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    This can't be right.

    Consider 1 spring.
    Energy conservation tells us that ##\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 = C##.
    This yields the period ##2\pi \sqrt{\frac m {D_1}}##.

    Now consider 2 opposite springs.
    A deviation in the direction of the springs gives:
    $$\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 + \frac 1 2 D_2 x^2 = C$$
    $$\frac 1 2 m \dot x^2 + \frac 1 2 (D_1 + D_2) x^2 = C$$
    This is equivalent to 1 spring with constant ##D_1 + D_2##.
    At least this goes into the direction of the answer.

    Now consider 3 springs in a equilateral triangle with equal spring constants ##D_1=D_2=D_3## and a small deviation into the the direction of the first spring:
    $$\frac 1 2 m \dot x^2 + \frac 1 2 D_1 x^2 + \frac 1 2 D_2 (\frac 1 2 x)^2 + \frac 1 2 D_3 (\frac 1 2 x)^2 = C$$
    $$\frac 1 2 m \dot x^2 + \frac 1 2 (D_1 + \frac 1 2 D_2 + \frac 1 2 D_3) x^2 = C$$
    The resulting period is equivalent to a spring with constant ##D_1 + \frac 1 2 D_2 + \frac 1 2 D_3## and not a spring constant of ##D_1 + D_2 + D_3##.

    The same principle applies to 4 springs in a tetrahedron.
    Last edited: Jan 9, 2014
  7. Jan 9, 2014 #6


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    Interesting. I find that the restoring force does not depend on the direction of the displacement from equilibrium and the restoring force obeys Hooke's law even for large displacements. The effective force constant has a simple relationship to the individual spring constants.
  8. Jan 9, 2014 #7


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    Consider the general expression for the potential energy of the system for an arbitrary location of the astronaut. Keep in mind that you are to neglect the natural length of each spring.
  9. Jan 9, 2014 #8


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    That would be true if the springs were initially at their relaxed lengths. Here, they're to be taken as effectively zero relaxed length.
  10. Jan 9, 2014 #9

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    Good point.
    The result for 2 springs remains the same though.
  11. Jan 9, 2014 #10


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    It's quite easy just using position and force vectors.
    If vi is the vector position of one vertex, x the vector of the astronaut at equilibrium position, and Δx a displacement vector, what is the vector for the force towards that vertex?
  12. Jan 9, 2014 #11


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    Yes! It does fall out almost immediately with forces and position and displacement vectors.

    I wanted to avoid having to think about vectors.
  13. Jan 10, 2014 #12
    Hi ILS and haruspex! :smile:

    What is wrong with ILS method? haruspex, I am unable to interpret "they're to be taken as effectively zero relaxed length.". :confused:

    The force vector towards the vertex is along Δx-x+vi, is this correct?

    If the above is correct, the force vector is:
    $$k\left(\left|\Delta \vec{x}-\vec{x}+\vec{v_i}\right|-\left|\vec{x}-\vec{v_i}\right|\right)\frac{\Delta \vec{x}-\vec{x}+\vec{v_i}}{\left|\Delta \vec{x}-\vec{x}+\vec{v_i}\right|}$$
    where k is the spring constant.
    Last edited: Jan 10, 2014
  14. Jan 10, 2014 #13
    Let the non-equilibrium position be ##\vec X##. Then the force due to the i-th spring must be directed from ##\vec X## to ##\vec v_i##. Is that ##\vec X - \vec v_i## or ##\vec v_i - \vec X##?

    Now back to the equilibrium position ##\vec x## and the displacement ##\Delta \vec x##. What is ##\vec X## in these?

    Note that the relaxed length is negligible. What is the elongation of the i-th spring in this case?
  15. Jan 10, 2014 #14
    It's the latter, sorry, I made a dumb mistake. :redface:
    ##\vec X## is ##\vec x + \vec {\Delta x}##.

    I am not getting this, why is the relaxed length negligible? :confused:
  16. Jan 10, 2014 #15
    Because the problem said so.
  17. Jan 10, 2014 #16

    The initial length of spring given by ##|\vec{x}-\vec{v_i}|## is negligible hence the extension in spring is given by ##|\vec{x}+\vec{\Delta x}-\vec{v_i}|##, correct?
  18. Jan 10, 2014 #17
    I am not sure what "initial" means here, but Hooke's law deals with the extension from the relaxed length. ##|\vec{x}-\vec{v_i}|## is the length at equilibrium, not the relaxed length.

  19. Jan 10, 2014 #18
    Do I have to write the total potential energy of the springs?

    The potential energy in the springs when the astronaut is displaced from equilibrium position is given by:
    $$\sum_{i=1}^4 \frac{1}{2}D_i |\vec{x}+\vec{\Delta x}-\vec{v_i}|^2$$
    Do I just expand the expression? :confused:
  20. Jan 10, 2014 #19
    As haruspex suggested, it is probably best to stay with forces. Find the resultant.
  21. Jan 10, 2014 #20
    The resultant force is given by:
    $$\vec{F}=\sum_{i=1}^4 D_i\vec{v_i}-\vec{x}\sum_{i=1}^4 D_i -\vec{\Delta x}\sum_{i=1}^4 D_i$$

    The above equation is of SHM with angular frequency ##\sqrt{D_1+D_2+D_3+D_4}##. This is the correct answer. Thanks a lot everyone for the help. :smile:

    So, there is no relation between this problem and the case of mass hanging by 4 parallel springs, correct?
  22. Jan 10, 2014 #21


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    and from post #12
    So putting those together, what is the force vector?
  23. Jan 10, 2014 #22
    It is instructive to look back at the problem and find out why the result is that way.

    Is it because of the tetrahedron?

    Is it because of the ISS environment?
  24. Jan 10, 2014 #23

    I really have no clue. :(
  25. Jan 10, 2014 #24
    Did you use the tetrahedron's symmetries in your derivation? Did it in fact matter that there were four springs and not twenty four?

    How did the ISS thing affect your derivation? What does that even mean? What if whatever that means was removed, would the result change?
  26. Jan 10, 2014 #25
    It did not affect the derivation.

    I am not sure, I guess that means we can make any arbitrary kind of connection of springs and the result (for time period) we get is always the same.
    The result won't change.

    Am I correct?
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